Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $k = \dfrac{q - 10}{-2q^2 + 10q + 100} \div \dfrac{q - 5}{-2q^2 - 28q - 90} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{q - 10}{-2q^2 + 10q + 100} \times \dfrac{-2q^2 - 28q - 90}{q - 5} $ First factor out any common factors. $k = \dfrac{q - 10}{-2(q^2 - 5q - 50)} \times \dfrac{-2(q^2 + 14q + 45)}{q - 5} $ Then factor the quadratic expressions. $k = \dfrac {q - 10} {-2(q + 5)(q - 10)} \times \dfrac {-2(q + 5)(q + 9)} {q - 5} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {(q - 10) \times -2(q + 5)(q + 9) } { -2(q + 5)(q - 10) \times (q - 5)} $ $k = \dfrac {-2(q + 5)(q + 9)(q - 10)} {-2(q + 5)(q - 10)(q - 5)} $ Notice that $(q + 5)$ and $(q - 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {-2\cancel{(q + 5)}(q + 9)(q - 10)} {-2\cancel{(q + 5)}(q - 10)(q - 5)} $ We are dividing by $q + 5$ , so $q + 5 \neq 0$ Therefore, $q \neq -5$ $k = \dfrac {-2\cancel{(q + 5)}(q + 9)\cancel{(q - 10)}} {-2\cancel{(q + 5)}\cancel{(q - 10)}(q - 5)} $ We are dividing by $q - 10$ , so $q - 10 \neq 0$ Therefore, $q \neq 10$ $k = \dfrac {-2(q + 9)} {-2(q - 5)} $ $ k = \dfrac{q + 9}{q - 5}; q \neq -5; q \neq 10 $